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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)
\(\Rightarrow n+1=50\)
\(\Rightarrow n=49\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)
\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)
\(\Rightarrow2n+1=51\)
\(\Rightarrow2n=50\)
\(\Rightarrow n=25\)
\(c)\)
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)
\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)
\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)
\(\Rightarrow x=\frac{199}{25}\)
Vậy \(x=\frac{199}{25}\)
~ Ủng hộ nhé
\(a)2.x-3=x+\frac{1}{2}\)
\(\Rightarrow2x-3-x=\frac{1}{2}\)
\(\Rightarrow x-3=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
\(b)4.x-\left(2.x+1\right)=3-\frac{1}{3}+x\)
\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)
\(\Rightarrow2.x-1=\frac{8}{3}+x\)
\(\Rightarrow2x-1-x=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}+1\)
\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)
\(\Rightarrow x=\frac{11}{3}\)
Vậy \(x=\frac{11}{3}\)
~ Ủng hộ nhé
\(S=1+5+5^2+5^4+...+5^{200}\)
\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S-S=5^{202}-1\)
\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)
a) S = 1 + 52 + 54 + ... + 5200
=> 52S = 52.(1 + 52 + 54 + ... + 5200)
=> 25S = 52 + 54 + 56 + ... + 5202
=> 25S - S = (52 + 54 + 56 + ... + 5202) - (1 + 52 + 54 + ... + 5200)
=> 24S = 5202 - 1
=> S = \(\frac{5^{202}-1}{24}\)
a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
Vậy ...
Bạn tham khảo nhé
a ) Ta có :
\(\left(-\frac{1}{5}\right)^{300}=\left(\frac{1}{5}\right)^{300}=\frac{1}{5^{300}}=\frac{1}{\left(5^3\right)^{100}}=\frac{1}{125^{100}}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(\frac{1}{3}\right)^{500}=\frac{1}{3^{500}}=\frac{1}{\left(3^5\right)^{100}}=\frac{1}{243^{100}}\)
Do \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\left(125^{100}< 243^{100}\right)\)
\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)
b )
Ta có :
\(2550^{10}=\left(50.51\right)^{10}=50^{10}.51^{10}\)
\(50^{20}=50^{10}.50^{10}\)
Do \(50^{10}.51^{10}>50^{10}.50^{10}\)
\(\Rightarrow50^{20}< 2550^{10}\)
c )
Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(5^{50}=\left(5^2\right)^{25}=25^{25}\)
Do \(16^{25}< 25^{25}< 27^{25}\)
\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)
1) So sánh
Ta có : 224 = 23.8 = (23)8 = 88
316 = 32.8 = (32)8 = 98
Vì 88 < 98
=> 224 < 316
2) Tính
\(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)^4.1024=\frac{1}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=\frac{2^{10}}{2^8}=2^2=4\)
3) Tìm x nguyên
(x - 1)x + 2 = (x - 1)x + 6
=> (x - 1)x + 6 - (x - 1)x + 2 = 0
=> (x - 1)x + 2.[(x - 1)4 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1^4\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 0 => x = 1(tm)
Nếu x - 1 = - 1 => x = 0(tm)
Nếu x - 1 = 1 => x = 2(tm)
Vậy \(x\in\left\{1;0;2\right\}\)
Bài 1:Ta có:
2^24=2^(6.4)=64^4
3^16=3^(4.4)=81^4
Bài 2.Ta có:
(0.25)^4=1/4.1/4.1/4.1/4=1/256
=>1/256.1024=4
Bài 3:
Ta có:(x-1)^(x+2)=(x-1)^(x+6)
Chia hai vế cho (x-1)^(x+2),do đó:
1=(x-1)^(x+4)
<=>x-1=1
<=>x=2
Hoặc chia hai vế cho (x-1)^(x+6)
(x-1)^(x-4)=1
<=>x-1=1
<=>x=2